Question

Solve each inequality algebraically.

$\frac{{(x+5)}^2}{x^2-4}\ge 0$

Short answer

Solution to the inequality $\frac{{(x+5)}^2}{x^2-4}\ge 0$is $(-\infty ,-2)\cup (2,\infty )$.

Step-by-step solution

Step 1. Given information 

We have been given an inequality $\frac{{(x+5)}^2}{x^2-4}\ge 0$.

We have to solve this inequality algebraically.

Step 2. Determine the real numbers at which the expression f equals zero and at which the expression f is undefined.  

Assume $f\left(x\right)=\frac{{(x+5)}^2}{x^2-4}$.

role="math" localid="1646126386160" $$\begin{gathered} {(x+5)}^2=0 \\ x+5=0 \\ x=-5 \end{gathered}$$

and

$$\begin{gathered} x^2-4=0 \\ (x+2)(x-2)=0 \\ x=-2,2 \end{gathered}$$

Step 3. Form the intervals  

Using the values of x found in previous step, we can divide the real numbers in the intervals:

$(-\infty ,-2)\cup (-2,2)\cup (2,\infty )$

Step 4. Select a number in each interval and evaluate f at the number  

Create the following table:

Interval	$(-\infty ,-2)$	$(-2,2)$	$(2,\infty )$
Number chosen	$-3$	0	3
Value of f	$f(-3)=\frac{4}{5}$	$f\left(0\right)=-\frac{25}{4}$	$f\left(3\right)=\frac{64}{5}$
Conclusion	positive	negative	positive

Step 5. Identify the interval 

Since we want to know where f is positive or zero, we conclude that $f\left(x\right)\ge 0$in the interval$(-\infty ,-2)\cup (2,\infty )$.